\(\int \frac {A+B x^2+C x^4}{(d+e x^2)^{5/2} (a+b x^2+c x^4)^2} \, dx\) [165]

Optimal result

Integrand size = 38, antiderivative size = 2548 \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{5/2} \left (a+b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Output:

1/6*e*(A*(3*b^3*d*e^2+3*b*c*d*(-3*a*e^2+c*d^2)+4*a*c*e*(-2*a*e^2+3*c*d^2)- 
b^2*(-2*a*e^3+6*c*d^2*e))-a*d*(B*(6*c^2*d^2+5*b^2*e^2-2*c*e*(7*a*e+3*b*d)) 
-C*(2*b^2*d*e-20*a*c*d*e+3*b*(a*e^2+c*d^2))))*x/a/(-4*a*c+b^2)/d/(a*e^2-b* 
d*e+c*d^2)^2/(e*x^2+d)^(3/2)-1/6*e*(A*(3*b^4*d^2*e^3-2*a*c*e*(-8*a^2*e^4-5 
9*a*c*d^2*e^2+9*c^2*d^4)-b*c*d*(64*a^2*e^4-27*a*c*d^2*e^2+3*c^2*d^4)-b^3*( 
-16*a*d*e^4+9*c*d^3*e^2)+b^2*(-4*a^2*e^5-40*a*c*d^2*e^3+9*c^2*d^4*e))+a*d* 
(B*(6*c^3*d^4-b^2*e^3*(2*a*e+13*b*d)-c^2*d^2*e*(106*a*e+9*b*d)+c*e^2*(8*a^ 
2*e^2+49*a*b*d*e+31*b^2*d^2))+C*d*(4*b^3*d*e^2+2*a*c*e*(-19*a*e^2+41*c*d^2 
)-b*c*d*(25*a*e^2+3*c*d^2)-b^2*(-11*a*e^3+16*c*d^2*e))))*x/a/(-4*a*c+b^2)/ 
d^2/(a*e^2-b*d*e+c*d^2)^3/(e*x^2+d)^(1/2)+1/2*x*(A*(3*a*b*c*e-2*a*c^2*d-b^ 
3*e+b^2*c*d)-a*(2*B*a*c*e-B*b^2*e+B*b*c*d+C*a*b*e-2*C*a*c*d)+c*(A*(2*a*c*e 
-b^2*e+b*c*d)-a*(-B*b*e+2*B*c*d+2*C*a*e-C*b*d))*x^2)/a/(-4*a*c+b^2)/(a*e^2 
-b*d*e+c*d^2)/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a)-1/2*c*(A*(b^5*d*e^3+b^3*e*(3 
*c^2*d^3-6*a*(-4*a*c+b^2)^(1/2)*e^3-c*d*e*(3*(-4*a*c+b^2)^(1/2)*d-8*a*e))- 
b^4*e^2*(3*c*d^2-e*((-4*a*c+b^2)^(1/2)*d-6*a*e))+2*a*c^2*(6*c^2*d^4-c*d^2* 
e*((-4*a*c+b^2)^(1/2)*d-30*a*e)-a*e^3*(21*(-4*a*c+b^2)^(1/2)*d+16*a*e))-b^ 
2*c*(c^2*d^4-3*c*d^2*e*((-4*a*c+b^2)^(1/2)*d+3*a*e)-2*a*e^3*(5*(-4*a*c+b^2 
)^(1/2)*d+17*a*e))+b*c*(22*a^2*(-4*a*c+b^2)^(1/2)*e^4+a*c*d*e^2*(3*(-4*a*c 
+b^2)^(1/2)*d-64*a*e)-c^2*d^3*((-4*a*c+b^2)^(1/2)*d+28*a*e)))+a*(b^4*B*d*e 
^3-b^3*e*(3*c*d^2*(B*e+2*C*d)-e^2*(B*(-4*a*c+b^2)^(1/2)*d-3*C*a*d+4*B*a...
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(88367\) vs. \(2(2548)=5096\).

Time = 23.82 (sec) , antiderivative size = 88367, normalized size of antiderivative = 34.68 \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{5/2} \left (a+b x^2+c x^4\right )^2} \, dx=\text {Result too large to show} \] Input:

Integrate[(A + B*x^2 + C*x^4)/((d + e*x^2)^(5/2)*(a + b*x^2 + c*x^4)^2),x]
 

Output:

Result too large to show
 

Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{5/2} \left (a+b x^2+c x^4\right )^2} \, dx\)

\(\Big \downarrow \) 2256

\(\displaystyle \int \left (\frac {-a C+A c+x^2 (B c-b C)}{c \left (d+e x^2\right )^{5/2} \left (a+b x^2+c x^4\right )^2}+\frac {C}{c \left (d+e x^2\right )^{5/2} \left (a+b x^2+c x^4\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(A c-a C) \int \frac {1}{\left (e x^2+d\right )^{5/2} \left (c x^4+b x^2+a\right )^2}dx}{c}+\frac {(B c-b C) \int \frac {x^2}{\left (e x^2+d\right )^{5/2} \left (c x^4+b x^2+a\right )^2}dx}{c}-\frac {2 c C \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {x \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}} \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )^{3/2} \left (a e^2-b d e+c d^2\right )}-\frac {2 c C \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \arctan \left (\frac {x \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {d+e x^2}}\right )}{\sqrt {\sqrt {b^2-4 a c}+b} \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )^{3/2} \left (a e^2-b d e+c d^2\right )}+\frac {C e x \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right )}{d \sqrt {d+e x^2} \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )}+\frac {C e x \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right )}{d \sqrt {d+e x^2} \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right ) \left (a e^2-b d e+c d^2\right )}+\frac {2 C e^2 x}{3 c d^2 \sqrt {d+e x^2} \left (a e^2-b d e+c d^2\right )}+\frac {C e^2 x}{3 c d \left (d+e x^2\right )^{3/2} \left (a e^2-b d e+c d^2\right )}\)

Input:

Int[(A + B*x^2 + C*x^4)/((d + e*x^2)^(5/2)*(a + b*x^2 + c*x^4)^2),x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2256

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^ 
(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4 
)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

Maple [F(-1)]

Timed out.

\[\int \frac {C \,x^{4}+B \,x^{2}+A}{\left (e \,x^{2}+d \right )^{\frac {5}{2}} \left (c \,x^{4}+b \,x^{2}+a \right )^{2}}d x\]

Input:

int((C*x^4+B*x^2+A)/(e*x^2+d)^(5/2)/(c*x^4+b*x^2+a)^2,x)
 

Output:

int((C*x^4+B*x^2+A)/(e*x^2+d)^(5/2)/(c*x^4+b*x^2+a)^2,x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{5/2} \left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate((C*x^4+B*x^2+A)/(e*x^2+d)^(5/2)/(c*x^4+b*x^2+a)^2,x, algorithm=" 
fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{5/2} \left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate((C*x**4+B*x**2+A)/(e*x**2+d)**(5/2)/(c*x**4+b*x**2+a)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{5/2} \left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {C x^{4} + B x^{2} + A}{{\left (c x^{4} + b x^{2} + a\right )}^{2} {\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((C*x^4+B*x^2+A)/(e*x^2+d)^(5/2)/(c*x^4+b*x^2+a)^2,x, algorithm=" 
maxima")
 

Output:

integrate((C*x^4 + B*x^2 + A)/((c*x^4 + b*x^2 + a)^2*(e*x^2 + d)^(5/2)), x 
)
 

Giac [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{5/2} \left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate((C*x^4+B*x^2+A)/(e*x^2+d)^(5/2)/(c*x^4+b*x^2+a)^2,x, algorithm=" 
giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{5/2} \left (a+b x^2+c x^4\right )^2} \, dx=\int \frac {C\,x^4+B\,x^2+A}{{\left (e\,x^2+d\right )}^{5/2}\,{\left (c\,x^4+b\,x^2+a\right )}^2} \,d x \] Input:

int((A + B*x^2 + C*x^4)/((d + e*x^2)^(5/2)*(a + b*x^2 + c*x^4)^2),x)
 

Output:

int((A + B*x^2 + C*x^4)/((d + e*x^2)^(5/2)*(a + b*x^2 + c*x^4)^2), x)
 

Reduce [F]

\[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{5/2} \left (a+b x^2+c x^4\right )^2} \, dx=\int \frac {1}{\sqrt {e \,x^{2}+d}\, a \,d^{2}+2 \sqrt {e \,x^{2}+d}\, a d e \,x^{2}+\sqrt {e \,x^{2}+d}\, a \,e^{2} x^{4}+\sqrt {e \,x^{2}+d}\, b \,d^{2} x^{2}+2 \sqrt {e \,x^{2}+d}\, b d e \,x^{4}+\sqrt {e \,x^{2}+d}\, b \,e^{2} x^{6}+\sqrt {e \,x^{2}+d}\, c \,d^{2} x^{4}+2 \sqrt {e \,x^{2}+d}\, c d e \,x^{6}+\sqrt {e \,x^{2}+d}\, c \,e^{2} x^{8}}d x \] Input:

int((C*x^4+B*x^2+A)/(e*x^2+d)^(5/2)/(c*x^4+b*x^2+a)^2,x)
 

Output:

int(1/(sqrt(d + e*x**2)*a*d**2 + 2*sqrt(d + e*x**2)*a*d*e*x**2 + sqrt(d + 
e*x**2)*a*e**2*x**4 + sqrt(d + e*x**2)*b*d**2*x**2 + 2*sqrt(d + e*x**2)*b* 
d*e*x**4 + sqrt(d + e*x**2)*b*e**2*x**6 + sqrt(d + e*x**2)*c*d**2*x**4 + 2 
*sqrt(d + e*x**2)*c*d*e*x**6 + sqrt(d + e*x**2)*c*e**2*x**8),x)